We divide the proof into four steps: the first two steps deal, respectively, with the first two points in the theorem, and the third point in the theorem is proved in the last two steps.

Step 1.
Since \(b\in L_2(\mathbb {R}^2)\)  implies \(b \in BMO^{-1}(\mathbb {R}^2)\) , we have thanks to Proposition REF  that
\(-\Delta \psi + u\cdot \nabla \psi = 0\quad \mbox{in }\mathbb {R}^2.\) 
By setting \(\psi ^R(x) := \psi (R x)\)  (\(R>0\) ), we see that \(\psi ^R\)  satisfies the conditions in Lemma REF , where we note that the maximum principle property is inherited from the previous drift-diffusion equation for \(\psi \) . Consequently, by taking \(r=1/2\)  in the oscillation lemma, we obtain:
\(\sup _{x\in B(1/2)}|\psi ^R(x) - \psi (0)| \le c\Vert \nabla \psi ^R\Vert _{L_2(B\setminus B(1/2))},\) 
thus
\(\sup _{x\in B(R/2)}|\psi (x) - \psi (0)| \le c\Vert b\Vert _{L_2(B(R)\setminus B(R/2))}\rightarrow 0\mbox{ as }R\rightarrow \infty .\) 
This implies that \(\psi \)  is constant and therefore \(b\equiv 0\) ; consequently we have that
\(-\Delta u + u\cdot \nabla u + \nabla \pi = 0,\quad \operatornamewithlimits{\mbox{div}}u = 0\quad \mbox{in }\mathbb {R}^2,\) 
with \(\nabla u\in L_2(\mathbb {R}^2)\) . This implies, thanks to Remark REF , that \(u\)  is constant.


Step 2.
By applying the divergence operator to the first equation in (REF ), we find that
\(-\Delta \pi = \operatornamewithlimits{\mbox{div}}\operatornamewithlimits{\mbox{div}}(u\otimes u - b\otimes b)\quad \mbox{in }\mathbb {R}^2;\) 
this guarantees, if we set \(r = \max (p,q)\) , that
\(\Vert \pi \Vert _{L_{\frac{r}{2}}(\mathbb {R}^2)} \le c\left( \Vert u\Vert ^2_{L_r(\mathbb {R}^2)} + \Vert b\Vert ^2_{L_r(\mathbb {R}^2)}\right);\) 
to obtain estimate (REF ), we used the hypothesis, the embedding of \(u,b\)  in \(BMO(\mathbb {R}^2)\)  (due to (REF )) and the interpolation between \(L_s\)  and \(BMO\)  spaces; see [1]}. A detailed justification of this estimate (in a more general context) is given later in the document (see `Step 1 (Pressure estimates)' of the proof of the first point in Theorem REF ).
Next, we rewrite the first equation in (REF ), in the following way:
\(-\Delta u + \nabla \left(\frac{|u|^2}{2} + \pi -\frac{|b|^2}{2}\right) -u^{\perp }\operatorname{curl}u = -b^{\perp }\operatorname{curl}b.\) 
By taking the scalar product with \(u\)  in the previous line, we get that
\(-\Delta \frac{|u|^2}{2} + |\nabla u|^2 + u\cdot \nabla \left(\frac{|u|^2}{2} + \pi -\frac{|b|^2}{2}\right) = -u\cdot b^{\perp }\operatorname{curl}b.\) 
Meanwhile, in view of the equality \(\nabla \psi = -b^{\perp }\)  and thanks to (REF ) (together with Remark REF ), we have that
\(\operatorname{curl}b = u\cdot b^{\perp }\quad \mbox{in }\mathbb {R}^2.\) 
Consequently, returning to the previous elliptic equation, we have that
\(|\nabla u|^2 + |\operatorname{curl}b|^2 = \Delta \frac{|u|^2}{2} - u\cdot \nabla \left(\frac{|u|^2}{2} + \pi -\frac{|b|^2}{2}\right)\quad \mbox{in }\mathbb {R}^2.\) 
For ease of notation, set \(Q := |u|^2/2 + \pi -|b|^2/2 \in L_{\frac{r}{2}}(\mathbb {R}^2)\) . By multiplying (REF ) by our usual rescaled cut-off function \(\varphi _R\)  and then integrating, we obtain that
\(\int _{B(R/2)}(|\nabla u|^2 + |\operatorname{curl}b|^2) \mathrm {\,d}x &\le \int _{B(R)\setminus B(R/2)}\frac{|u|^2}{2}\Delta \varphi _R \mathrm {\,d}x + \int _{B(R)\setminus B(R/2)} u\cdot \nabla \varphi _R Q \mathrm {\,d}x\\&\le \frac{c(\varphi ,p)}{R^{\frac{4}{p}}}\left(\int _{B(R)\setminus B(R/2)}|u|^p \mathrm {\,d}x\right)^{\frac{2}{p}} + \\& + \frac{c(\varphi ,p,q)}{R^{1-\frac{2}{s}}}\left(\int _{B(R)\setminus B(R/2)}|u|^p \mathrm {\,d}x\right)^{\frac{1}{p}}\left(\int _{B(R)\setminus B(R/2)}|Q|^{\frac{r}{2}} \mathrm {\,d}x\right)^{\frac{2}{r}},\) 

with \(1/s = 1 - 1/p - 2/r\) . Thus, the right-hand side in last inequality above vanishes in the limit as \(R\rightarrow \infty \)  whenever \(1 - 2/s\ge 0\) , i.e., \(1/p + 2/r \ge 1/2\) . And the second point is proved.


Step 3.
In this penultimate step, we prove the third point in the theorem for the non-critical cases \(p \in ]1,2[\) . Firstly, observe that, for this range of exponents, the div-curl lemma of Coifman, Lions, Meyer and Semmes (cf. [2]}) applies. Due to the incompressibility condition, we deduce that \(u \cdot \nabla \psi \)  belongs to the Hardy space \(\mathcal {H}^1(\mathbb {R}^2)\) . Now, define
\(v := (-\Delta )^{-1} (-u \cdot \nabla \psi ),\) 
i.e.,
\(v(x) = \frac{1}{2\pi }\int _{\mathbb {R}^2} \log |x-y| (u\cdot \nabla \psi )(y) \, \mathrm {d}y, \qquad \forall x \in \mathbb {R}^2.\) 
Observe that the above integral is well-defined since the function \(y \mapsto \log |y|\)  belongs to the space \(BMO(\mathbb {R}^2)\) , which is the dual of \(\mathcal {H}^1(\mathbb {R}^2)\) ; in fact, we readily deduce that \(v \in L_\infty (\mathbb {R}^2)\) . Moreover,
an application of Lebesgue's dominated convergence theorem shows that
\(\nabla v(x) = \frac{1}{2\pi } \int _{\mathbb {R}^2} \frac{(x-y)}{|x-y|^2} (u \cdot \nabla \psi )(y) \, \mathrm {d}y, \qquad \forall x \in \mathbb {R}^2.\) 
The above and, for instance, [3]} and [4]}, yield \(\nabla v \in L_2(\mathbb {R}^2)\) . Additionally, using the fundamental property of the Poisson kernel in (REF ), we obtain, in the sense of distributions,
\(-\Delta v + u \cdot \nabla \psi = 0 \qquad \text{in } \mathbb {R}^2.\) 
Our goal is now to “transfer” the square-integrability of \(\nabla v\)  onto \(\nabla \psi \) , since these satisfy similar equations. To this end, returning to the drift-diffusion equation for the stream function \(\psi \)  and taking an additional derivative, we find that
\(-\Delta (\nabla \psi - \nabla v) = 0 \qquad \text{in } \mathbb {R}^2.\) 
Since \(\nabla \psi - \nabla v \in L_{p^{\prime }}(\mathbb {R}^2) + L_2(\mathbb {R}^2)\) ,
we deduce that \(\nabla \psi - \nabla v = 0\)  in \(\mathbb {R}^2\) . Hence, we get that \(b \in L_2(\mathbb {R}^2)\)  and conclude as we did in `Step 1' of this proof.


Step 4.
Finally, we prove the third point in the theorem for the critical case \(p=1\) . In this case, we cannot apply the div-curl lemma of Coifman, Lions, Meyer and Semmes. Nevertheless, we can still construct a solution of the problem \(-\Delta v + u \cdot \nabla \psi = 0\)  with the desired properties, i.e., \(v \in L_\infty (\mathbb {R}^2)\)  with \(\nabla v \in L_2(\mathbb {R}^2)\) . This is encapsulated in Lemma REF , the proof of which is delayed until the Appendix. Comparing the equation for \(v\)  with the drift-diffusion equation for the stream function \(\psi \) , we find once again
\(-\Delta (\psi - v) = 0 \qquad \text{in } \mathbb {R}^2.\) 
By applying the operator \(\nabla ^{\perp }\)  to the previous equation and by noticing that \(b - \nabla ^{\perp } v \in L_{\infty }(\mathbb {R}^2) + L_2(\mathbb {R}^2)\) , we obtain that
there exists a constant \(a \in \mathbb {R}^2\)  such that
\(b = a + \nabla ^{\perp } v \qquad \mbox{in } \mathbb {R}^2.\) 
From the previous identity, we also get that \(\nabla v \in L_2(\mathbb {R}^2)\cap L_{\infty }(\mathbb {R}^2)\) , which leads,
by interpolation, to
\(\nabla v \in L_q(\mathbb {R}^2), \qquad \forall q \in [2,\infty ].\) 
We find, in this case, the following equation for the pressure:
\(-\Delta \pi = \operatornamewithlimits{\mbox{div}}\operatornamewithlimits{\mbox{div}}(u\otimes u - a\otimes \nabla ^\perp v - \nabla ^\perp v\otimes a - \nabla ^\perp v\otimes \nabla ^\perp v),\) 
which leads, without loss of generality, to
\(\Vert \pi \Vert _{L_r(\mathbb {R}^2)} \le c\Vert u\otimes u - a\otimes \nabla ^\perp v - \nabla ^\perp v\otimes a - \nabla ^\perp v\otimes \nabla ^\perp v\Vert _{L_r(\mathbb {R}^2)},\) 
for all \(r\in [2,\infty [\)  (more details can be found in `Step 1 (Pressure estimates)' of the proof of the first point in Theorem REF ).
We can see that identity (REF ) still holds true in our current setting; notice now that
\(\nabla \frac{|b|^2}{2} = \nabla \frac{|\nabla v|^2}{2} + \nabla (a\cdot \nabla ^\perp v).\) 
Thus, by setting \(\tilde{Q} := |u|^2/2 + \pi - |\nabla v|^2/2 - a\cdot \nabla ^\perp v \in L_r(\mathbb {R}^2)\)  with \(r \in [2,\infty [\) , we have that
\(|\nabla u|^2 + |\operatorname{curl}b|^2 = \Delta \frac{|u|^2}{2} - u\cdot \nabla \tilde{Q} \quad \mbox{in }\mathbb {R}^2.\) 
Consequently, by testing the previous equation against our usual rescaled cut-off function \(\varphi _R\) , we obtain
\(\int _{B(R/2)}(|\nabla u|^2 + |\operatorname{curl}b|^2) \mathrm {\,d}x &\le \int _{B(R)\setminus B(R/2)}\frac{|u|^2}{2}\Delta \varphi _R \mathrm {\,d}x + \int _{B(R)\setminus B(R/2)} u\cdot \nabla \varphi _R \tilde{Q} \mathrm {\,d}x\\&\le \frac{c(\varphi )}{R^{2}} \int _{B(R)\setminus B(R/2)}|u|^2 \mathrm {\,d}x + \\& + \frac{c(\varphi )}{R}\left(\int _{B(R)\setminus B(R/2)}|u|^2 \mathrm {\,d}x\right)^{\frac{1}{2}}\left(\int _{B(R)\setminus B(R/2)}|\tilde{Q}|^2 \mathrm {\,d}x\right)^{\frac{1}{2}},\\& \rightarrow 0\quad \mbox{ as }R\rightarrow \infty .\) 

This proves the final point of the theorem.
Let us give now the proof of Theorem REF . We start with the second assertion in the theorem.
Because \(u,b\in BMO^{-1}(\mathbb {R}^2)\) , we have
that \(u = \nabla ^{\perp }\phi \)  and \(b = \nabla ^{\perp } \psi \)  with \(\phi ,\psi \in BMO(\mathbb {R}^2)\) . Thanks, to Proposition REF , we know that
\(-\Delta \psi + u\cdot \nabla \psi = 0;\) 
by multiplying the above equation by \((\psi - [\psi ]_{,R})\varphi _R^2\)  (where \(\varphi _R\)  is our usual rescaled cut-off function), and then integrating by parts, we get that
\(\int _{B(R)}|\nabla \psi |^2 \varphi _R^2 \mathrm {\,d}x = \frac{1}{2}\int _{B(R)}|\psi - [\psi ]_{,R}|^2\Delta \varphi ^2_R \mathrm {\,d}x + 2\int _{B(R)}\varphi _R\nabla \psi \cdot \nabla ^{\perp }\varphi _R (\psi - [\psi ]_{,R})(\phi - [\phi ]_{,R})\mathrm {\,d}x;\) 
this implies, thanks to Young's inequality,
\(\int _{B(R/2)}|\nabla \psi |^2 \mathrm {\,d}x &\le c(\varphi )\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\) 



B(R)|- [],R|2  dx + c()R2B(R)|- [],R|2|- [],R|2  dx
c()(2BMO(R2) + 2BMO(R2)2BMO(R2)).

Thus, by taking the limit \(R\rightarrow \infty \) , we get that \(\nabla \psi \in L_2(\mathbb {R}^2)\) , i.e., \(b\in L_2(\mathbb {R}^2)\) . By applying the first point of Theorem REF , we conclude the proof.
We prove now the first assertion in Theorem  REF
We divide the proof into two steps. Let us also recall that, since \(u \in BMO^{-1}(\mathbb {R}^2)\) , we have that \(u = \nabla ^{\perp } \phi \)  with \(\phi \in BMO(\mathbb {R}^2)\) .

Step 1 (Pressure estimates).
Let us start by noticing that, under our hypothesis, we have that:
\(-\Delta u + u\cdot \nabla u \in (\mathring{L}^1_2(\mathbb {R}^2))^*\) 
the dual of \(\mathring{L}^1_2(\mathbb {R}^2)\)  (the closure of \(C^{\infty }_0(\mathbb {R}^2)\)  with respect of the semi-norm \(\Vert \nabla \cdot \Vert _{L_2(\mathbb {R}^2)}\) ). To see this, consider \(\varphi \in C^{\infty }_0(\mathbb {R}^2)\) ; then, we have that
\(\langle -\Delta u_i + u\cdot \nabla u_i,\varphi \rangle &= \int _{\mathbb {R}^2}\nabla u_i \cdot \nabla \varphi \mathrm {\,d}x - \int _{\mathbb {R}^2} \varphi \nabla u_i\cdot \nabla ^{\perp } \varphi \mathrm {\,d}x\\&\le c\left(\Vert \nabla u\Vert _{L_2(\mathbb {R}^2)} + \Vert \varphi \Vert _{BMO(\mathbb {R}^2)}\Vert \nabla u\Vert _{L_2(\mathbb {R}^2)}\right)\Vert \nabla \varphi \Vert _{L_2(\mathbb {R}^2)}\quad (\forall i = 1,2),\) 

where, for the last inequality, we used the duality between \(BMO\)  and Hardy \(\mathcal {H}^1\)  spaces, and the div-curl lemma of Coifman, Lions, Meyer and Semmes (see for instance [2]}). Consequently, we have that (see for instance [6]}) there exists \(F\in L_2(\mathbb {R}^2;\mathbb {R}^{2\times 2})\)  such that
\(-\Delta u + u\cdot \nabla u = \operatornamewithlimits{\mbox{div}}(F);\) 
in particular, we deduce that
\(\nabla \pi \in (\mathring{L}^1_2(\mathbb {R}^2))^* + L_s(\mathbb {R}^2),\) 
with \(1/s = 1/q + 1/2\) . Next, let us introduce \(\pi _1\)  and \(\pi _2\)  defined as follow:
\(\pi _1 := (-\Delta )^{-1}\operatornamewithlimits{\mbox{div}}\operatornamewithlimits{\mbox{div}}(F)\quad \mbox{and}\quad \pi _2 := -(-\Delta )^{-1}\operatornamewithlimits{\mbox{div}}\operatornamewithlimits{\mbox{div}}(b\otimes b),\) 
where \(F\)  is as in (REF ).
We have that \(\pi _1 \in L_2(\mathbb {R}^2)\)  and \(\pi _2\in L_{\frac{q}{2}}(\mathbb {R}^2)\)  if \(q\ne \infty \)  or \(\pi _2 \in BMO(\mathbb {R}^2)\)  if \(q = \infty \) , and \(\nabla \pi _2 \in L_s(\mathbb {R}^2)\)  (with \(1/s = 1/q + 1/2\) ). Going back to the first equation in (REF ), we get that
\(-\Delta (\pi - \pi _1 - \pi _2) = 0;\) 
but since \(\nabla (\pi - \pi _1 - \pi _2) \in (\mathring{L}^1_2(\mathbb {R}^2))^* + L_s(\mathbb {R}^2)\)  (see (REF ) and definition of \(\pi _1\)  and \(\pi _2\) ), we have that \(\nabla (\pi - \pi _1 - \pi _2) \equiv 0\) ; therefore, without loss of generality
\(\pi = \pi _1 + \pi _2.\) 


Step 2.
Similarly to what we did in `Step 2' of the proof of Theorem REF , we obtain that (see (REF )):
\(|\nabla u|^2 + |\operatorname{curl}b|^2 = \Delta \frac{|u|^2}{2} - u\cdot \nabla \left(\frac{|u|^2}{2} + \pi _1 + \pi _2 -\frac{|b|^2}{2}\right)\quad \mbox{in }\mathbb {R}^2.\) 
Before going any further, let us introduce the following special mean function:
\([u]_{\varphi ,R} := \int _{B(R)}u(x)\varphi _R(x) \mathrm {\,d}x\left( \int _{B(R)}\varphi _R(x)\mathrm {\,d}x \right)^{-1},\) 
where \(\varphi _R\)  is our usual rescaled cut-off function. Now, we multiply \(\varphi _R\)  and integrate over the ball \(B(R)\)  to obtain
\(&\int _{B(R/2)}\left( |\nabla u|^2 + |\operatorname{curl}b|^2\right)\mathrm {\,d}x \le \frac{1}{2}\int _{B(R)\setminus B(R/2)}|u|^2\Delta \varphi _R \mathrm {\,d}x\\& + \int _{B(R)\setminus B(R/2)}(\phi - [\phi ]_{,R})\nabla (|u|^2/2)\cdot \nabla ^{\perp } \varphi _R \mathrm {\,d}x + \int _{B(R)\setminus B(R/2)} \pi _1 u\cdot \nabla \varphi _R \mathrm {\,d}x\\& + \int _{B(R)\setminus B(R/2)}(\phi - [\phi ]_{,R})\nabla \pi _2\cdot \nabla ^{\perp } \varphi _R \mathrm {\,d}x + \int _{B(R)\setminus B(R/2)}(\phi - [\phi ]_{,R})\nabla (|b|^2/2)\cdot \nabla ^{\perp } \varphi _R \mathrm {\,d}x\\& =: \sum _{i = 1}^5 I_i(R).\) 

Our goal now is to estimate the \(I_i\)  terms. We have
\(I_1(R) &= \frac{1}{2}\int _{B(R)\setminus B(R/2)}| u - [u]_{B(R)\setminus B(R/2)}|^2\Delta \varphi _R \mathrm {\,d}x\\&\phantom{=} + [u]_{B(R)\setminus B(R/2)}\cdot \int _{B(R)\setminus B(R/2)}(u - [u]_{B(R)\setminus B(R/2)})\Delta \varphi _R \mathrm {\,d}x, \\&\le \frac{c(\varepsilon ,\varphi )}{R^2}\int _{B(R)\setminus B(R/2)}| u - [u]_{B(R)\setminus B(R/2)}|^2 \mathrm {\,d}x + 2\varepsilon (\left|[u]_{B(R)\setminus B(R/2)} - [u]_{\varphi ,R}\right|^2 + \left|[u]_{\varphi ,R}\right|^2)\\&\quad \le c(\varepsilon ,\varphi )\int _{B(R)\setminus B(R/2)}|\nabla u|^2 \mathrm {\,d}x + \varepsilon c(\varphi ) \int _{B(R)}|\nabla u|^2 \mathrm {\,d}x + \varepsilon \frac{c(\varphi )}{R^2}\Vert \phi \Vert ^2_{BMO(\mathbb {R}^2)}.\) 

Let us also point out that, for this last inequality, we used
\(\begin{aligned}\left|[u]_{B(R)\setminus B(R/2)} - [u]_{\varphi ,R}\right|^2 &\le c(\varphi )\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\end{aligned}\) 



B(R)|u - [u]B(R)B(R/2)|2  dx
c()B(R)|u|2  dx;


and we bound the term \(|[u]_{\varphi ,R}|\)  as follows:
\(|[u]_{\varphi ,R}| &= \left|\frac{1}{R^2\int _B \varphi (x)\mathrm {\,d}x}\int _{B(R)}\nabla ^{\perp }(\phi - [\phi ]_{,R})\varphi _R \mathrm {\,d}x\right|\\&\le \frac{c(\varphi )}{R}\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\) 



B(R)|- [],R| dx
c()RBMO(R2).
Next, we have
\(I_2(R) &= \int _{B(R)\setminus B(R/2)}(\phi - [\phi ]_{,R})\nabla (|u - [u]_{B(R)\setminus B(R/2)}|^2/2)\cdot \nabla \varphi _R \mathrm {\,d}x\\& \phantom{=} + [u]_{B(R)\setminus B(R/2)}\cdot \int _{B(R)\setminus B(R/2)}(\phi - [\phi ]_{,R})\nabla u\nabla ^{\perp }\varphi _R \mathrm {\,d}x\\& \le c(\varphi )\left(\int _{B(R)\setminus B(R/2)}|\nabla u|^2 \mathrm {\,d}x\right)^{\frac{1}{2}}\left(\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\right.\) 



B(R)|- [],R|4  dx)14
=(



B(R)B(R/2)|u - [u]B(R)B(B/2)|4  dx )14
= + c()|[u]B(R)B(B/2)|(B(R)B(R/2)|u|2  dx)12(



B(R)|- [],R|2  dx)12
 c()BMO(R2)B(R)B(R/2)|u|2  dx
= + c()BMO(R2)|[u]B(R)B(B/2)|(B(R)B(R/2)|u|2  dx)12;

but as before, we deal with the term \(\left|[u]_{B(R)\setminus B(B/2)}\right|\)  in the following manner:
\(\begin{split}\left|[u]_{B(R)\setminus B(B/2)}\right| &\le \left|[u]_{B(R)\setminus B(R/2)} - [u]_{\varphi ,R}\right| + \left|[u]_{\varphi ,R}\right|\\&\le c(\varphi )\left(\int _{B(R)}|\nabla u|^2 \mathrm {\,d}x \right)^{\frac{1}{2}} + \frac{c(\varphi )}{R}\Vert \phi \Vert _{BMO(\mathbb {R}^2)}.\end{split}\) 

Consequently, we have
\(I_2(R) \rightarrow 0\quad \mbox{ as }R\rightarrow \infty .\) 
Next, we estimate
\(I_3(R) &\le c(\varphi )\left(\int _{B(R)\setminus B(R/2)}|\pi _1|^2 \mathrm {\,d}x \right)^{\frac{1}{2}}\left( \mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\right.\) 



B(R)B(R/2)|u - [u]B(R)B(R/2)|2  dx)12
= + c()|[u]B(R)B(R/2)|(B(R)B(R/2)|1|2  dx )12;

therefore, because \(\pi _1 \in L_2(\mathbb {R}^2)\)  and thanks to (REF ), we get
\(I_3(R) \rightarrow 0\quad \mbox{ as }R\rightarrow \infty .\) 
Finally, since \(\nabla \pi _2, \nabla (|b|^2/2) \in L_s(\mathbb {R}^2)\)  with \(1/s = 1/q + 1/2\) , we have that \(I_4(R)\)  and \(I_5(R)\)  can be treated in the same way:
\(I_4(R) &\le \frac{c(\varphi ,q)}{R^{\frac{2}{q}}}\left(\int _{B(R)\setminus B(R/2)} |\nabla \pi _2|^s\mathrm {\,d}x \right)^{\frac{1}{s}}\left(\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\right.\) 



B(R)|- [],R|s' dx )1s'
= 0   as R.

Summarising our efforts, we obtain that
\(\int _{\mathbb {R}^2}\left(|\nabla u|^2 + |\operatorname{curl}b|^2\right)\mathrm {\,d}x \le \varepsilon c(\varphi )\int _{\mathbb {R}^2}|\nabla u|^2 \mathrm {\,d}x,\) 
for all \(\varepsilon >0\) ; and the proof is done.

Concluding remarks
At this point in time, it is still unclear to the authors whether one has a Liouville-type theorem for system (REF ) assuming only condition (REF ) to hold or, as suggested by Proposition REF , assuming in addition to condition (REF ) that either \(u\in BMO^{-1}(\mathbb {R}^2)\)  or \(b\in BMO^{-1}(\mathbb {R}^2)\) . This will be the object of future investigations.

appendix

Technical lemmas
For the reader's convenience, we present three technical results needed in the proof our main results.
First, we recall an inequality from [7]}.
Lemma 1.1 
Let \(f\in L_{2,loc}(\mathbb {R}^2)\)  with finite Dirichlet energy
\(\int _{\mathbb {R}^2}|\nabla f|^2 \mathrm {\,d}x <\infty .\) 
Then
\(\int _{B(R)}|f|^m \mathrm {\,d}x \le c R^2(\log 2R)^{\frac{m}{2}}\left( \int _{\mathbb {R}^2}|\nabla f|^2\mathrm {\,d}x + \int _{B(1)}|f|^2\mathrm {\,d}x\right)^{\frac{m}{2}},\) 
for all \(m,R\ge 1\)  and \(c>0\)  and absolute constant.

Secondly, we state an oscillation lemma inspired from [8]}; although our statement of this result is slightly different, the idea of the proof is the same. With the aim of keeping the paper self-contained, we also provide a proof.
Lemma 1.2 (Oscillation lemma)  
Let \(v\in C(\overline{B})\cap H^1(B)\)  such that for any \(r\in ]0,1[\)  the following maximum principle holds in \(B(r)\) :
\(\max _{B(r)} v = \max _{\partial B(r)} v, \quad \min _{B(r)} v = \min _{\partial B(r)} v.\) 
Then,
\(\sup _{x\in B(r)}|v(x) - v(0)| \le \frac{c}{\sqrt{-\log r}}\Vert \nabla v\Vert _{L_2(B\setminus B(r))},\) 
for all \(r\in ]0,1[\)  and with \(c>0\)  an absolute constant.

For \(r \in ]0,1[\) , using radial and angular coordinates, we write
\(\begin{split}\int _{B \setminus B(r)} |\nabla v|^2 \mathrm {\,d}x &= \int _r^{1}\int _0^{2\pi }|\nabla v(s,\theta )|^2 s \mathrm {\,d}\theta \mathrm {\,d}s \\&= \int _{r}^1 \int _0^{2\pi } \left[|\partial _s v|^2 + \frac{1}{s^2}|\partial _\theta v|^2\right] s \mathrm {\,d}\theta \mathrm {\,d}s \\&\ge \int _r^1 \frac{1}{s} \int _0^{2\pi } |\partial _\theta v(s,\theta )|^2 \mathrm {\,d}\theta \mathrm {\,d}s.\end{split}\) 
On the other hand, by the fundamental theorem of calculus,
\(v(s,\theta _1) - v(s,\theta _2) = \int _{\theta _1}^{\theta _2} \partial _\theta v(s,\theta ) \mathrm {\,d}\theta ,\) 

which implies (using the Cauchy-Schwarz inequality)
\(|v(s,\theta _1) - v(s,\theta _2)| &\le \int _{0}^{2\pi } |\partial _\theta v(s,\theta )| \mathrm {\,d}\theta \\&\le \sqrt{2\pi } \left(\int _0^{2\pi } |\partial _\theta v(s,\theta )|^2 \mathrm {\,d}\theta \right)^{1/2}.\) 

Hence
\(\max _{\theta _1 \in [0,2\pi [} v(s,\theta _1) - \min _{\theta _2 \in [0,2\pi [} v(s,\theta _2) &\le \sqrt{2\pi }\left(\int _0^{2\pi } |\partial _\theta v(s,\theta )|^2 \mathrm {\,d}\theta \right)^{1/2},\) 

which yields
\(\mathrm {osc}_{\partial B(s)} v&\le \sqrt{2\pi }\left(\int _0^{2\pi } |\partial _\theta v(s,\theta )|^2 \mathrm {\,d}\theta \right)^{1/2}.\) 

Using (REF ), we deduce
\(\int _r^1 (\mathrm {osc}_{\partial B(s)} v)^2 \frac{\mathrm {\,d}s}{2\pi s} \le \int _{B \setminus B(r)} |\nabla v|^2 \mathrm {\,d}x.\) 

We observe that \(\mathrm {osc}_{\partial B(s)} v \ge \mathrm {osc}_{\partial B(r)}v \ge 0\)  for \(s \in [r,1]\) . As a result,
\(\int _{B \setminus B(r)} |\nabla v|^2 \mathrm {\,d}x &\ge \frac{1}{2\pi }\int _r^1 \frac{1}{s} (\mathrm {osc}_{\partial B(s)} v)^2 \mathrm {\,d}s \\&\ge \frac{(\mathrm {osc}_{\partial B(r)} v)^2}{2\pi }\int _r^1 \frac{1}{s} \mathrm {\,d}s\\&= \frac{1}{2\pi } (-\log r)(\mathrm {osc}_{\partial B(r)} v)^2.\) 

Once again, by the maximum principle, we have that \(\mathrm {osc}_{B(r)} v = \mathrm {osc}_{\partial B(r)} v\) , so
\(\mathrm {osc}_{B(r)} v\le \frac{\sqrt{2\pi }}{\sqrt{-\log r}}\Vert \nabla v\Vert _{L_2(B\setminus B(r))},\) 

which concludes the proof.
Finally, we provide a technical lemma which was needed in the proof of Theorem REF .
Lemma 1.3 
Suppose \(\psi _1,\psi _2\)  are smooth functions such that \(\nabla \psi _1 \in L_1(\mathbb {R}^2)\)  and \(\nabla \psi _2 \in L_\infty (\mathbb {R}^2)\) . Then there exists a distributional solution \(v \in L_\infty (\mathbb {R}^2)\)  with \(\nabla v\in L_2(\mathbb {R}^2)\)  to the equation
\(-\Delta v = \nabla ^\perp \psi _1 \cdot \nabla \psi _2 \qquad \text{in } \mathbb {R}^2.\) 
Moreover, there exists a positive constant \(c\)  such that
\(\Vert v \Vert _{L_\infty (\mathbb {R}^2)} + \Vert \nabla v \Vert _{L_2(\mathbb {R}^2)} \le c \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)} \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)}.\) 

Recall our usual cut-off function \(\varphi \in C^{\infty }_0(\mathbb {R}^2)\)  such that \(0\le \varphi \le 1\)  with \(\varphi \equiv 1\)  in \(B(1/2)\)  and \(\varphi \equiv 0\)  in \(B\setminus B(3/4)\) . Now define, for each \(n\in \mathbb {N}\) , the rescaled cut-off function \(\varphi _n(x) := \varphi (x/n)\) , and
\(\hat{\psi }^n_1 := (\psi _1 - [\psi _1]_{,n}) \varphi _n , \qquad \hat{\psi }^n_2 := (\psi _2 - [\psi _2]_{,n}) \varphi _n.\) 
We have that
\(v^n := (-\Delta )^{-1}(\nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2)\) 
is well-defined and
\(-\Delta v^n = \nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2 \qquad \text{in } \mathbb {R}^2.\) 
Our aim is now to obtain estimates on \(v^n\)  that are independent of \(n\) . We begin by observing that
\(\begin{aligned}\Vert \nabla \hat{\psi }^n_1 \Vert _{L_1(\mathbb {R}^2)} &\le \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)} + \frac{c(\varphi )}{n} \int _{B(n)} |\psi _1 - [\psi _1]_{,n}| \, \mathrm {d}x \\&\le c(\varphi ) \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)},\end{aligned}\) 
where we used the triangle inequality and the boundedness of \(\varphi \)  and its derivative to obtain the first inequality, and the Poincaré–Wirtinger inequality on a ball to obtain the second inequality. Similarly, we have that
\(\Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \le \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)} + \frac{c(\varphi )}{n}\sup _{x \in B(n)} \bigg | \psi _2(x) - \mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\) 



B(n) 2(y)   dy |,

where we used once again the triangle inequality and the boundedness of \(\varphi \)  and its derivative. Then, due to the bound on the gradient of \(\psi _2\) , we have that
\(\frac{1}{n}\sup _{x \in B(n)} \bigg | \psi _2(x) - \frac{1}{|B(n)|}\int _{B(n)} \psi _2(y) \, \mathrm {d}y \bigg | &\le \frac{1}{n}\sup _{x\in B(n)}\mathchoice{{\displaystyle {\textstyle -}{\int } }\hbox{$\textstyle -$ }}{\hspace{0.0pt}}{-}{.}6\) 



B(n)|2 (x) - 2(y)| dy
c2L(R2),       n N,

whence,
\(\Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \le c(\varphi ) \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)}, \qquad \forall n \in \mathbb {N}.\) 
Hence, the sequence \(\lbrace \nabla \hat{\psi }^n_1\rbrace _{n\in \mathbb {N}}\)  is uniformly bounded in \(L_1(\mathbb {R}^2)\) , and \(\lbrace \nabla \hat{\psi }^n_2\rbrace _{n\in \mathbb {N}}\)  is uniformly bounded in \(L_\infty (\mathbb {R}^2)\) .
In what follows, we directly estimate \(v^n\)  using its singular integral representation; these calculations are inspired from similar estimates due to Wente in [9]}. By rewriting the right-hand side of (REF ) in polar coordinates \((s,\theta )\) , we find that
\(\nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2 = \frac{1}{s}\partial _s (\hat{\psi }^n_1 \partial _\theta \hat{\psi }^n_2) - \frac{1}{s}\partial _\theta (\hat{\psi }^n_1 \partial _s \hat{\psi }^n_2),\) 
and hence, using the periodicity with respect to the angular variable,
we get that the expression for \(v^n(0)\)  reads as
\(v^n(0) = - \frac{1}{2\pi } \int _0^{2\pi } \bigg (\int _{0}^\infty \log s~ \partial _s (\hat{\psi }^n_1 \partial _\theta \hat{\psi }^n_2) \, \mathrm {d}s \bigg ) \, \mathrm {d}\theta ,\) 
where we also applied the Fubini-Tonelli theorem to change the order of integration. In turn, after an integration by parts, this becomes
\(v^n(0) = \frac{1}{2\pi } \int _0^{2\pi } \bigg (\int _{0}^\infty \hat{\psi }^n_1 \frac{1}{s} \partial _\theta \hat{\psi }^n_2 \, \mathrm {d}s \bigg ) \, \mathrm {d}\theta ;\) 
note that the boundary terms have vanished since
\(|\log {s}~ \hat{\psi }^n_1 \partial _\theta \hat{\psi }^n_2| \le \Vert \hat{\psi }^n_1\Vert _{L_{\infty }(\mathbb {R}^2)} \Vert \nabla \hat{\psi }^n_2\Vert _{L_{\infty }(\mathbb {R}^2)} |s \log s|\rightarrow 0\qquad \mbox{as }s\rightarrow 0^+,\) 
and because \(\hat{\psi }^n_1\)  and \(\hat{\psi }^n_2\)  are compactly supported. Using once again the Fubini-Tonelli theorem and the periodicity with respect to the angular variable in the \(\partial _\theta \hat{\psi }^n_2\)  portion of the integrand, the previous expression may be rewritten as
\(v^n(0) = \frac{1}{2\pi } \int _{0}^\infty \frac{1}{s} \bigg [\int _0^{2\pi } \big ( \hat{\psi }^n_1 - \frac{1}{2\pi }\int _0^{2\pi } \hat{\psi }^n_1(s,\alpha ) \, \mathrm {d}\alpha \big )\partial _\theta \hat{\psi }^n_2 \, \mathrm {d}\theta \bigg ] \, \mathrm {d}s.\) 
In turn,
\(\begin{aligned}|v^n(0)| &\le \frac{1}{2\pi } \Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \int _{0}^\infty \int _0^{2\pi } \bigg | \hat{\psi }^n_1 - \frac{1}{2\pi }\int _0^{2\pi } \hat{\psi }^n_1(s,\alpha ) \, \mathrm {d}\alpha \bigg | \, \mathrm {d}s \, \mathrm {d}\theta \\&\le c \Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \int _{0}^\infty \int _0^{2\pi } | \partial _\theta \hat{\psi }^n_1| \, \mathrm {d}s \, \mathrm {d}\theta \\&\le c \Vert \nabla \hat{\psi }^n_2 \Vert _{L_\infty (\mathbb {R}^2)} \Vert \nabla \hat{\psi }^n_1 \Vert _{L_1(\mathbb {R}^2)},\end{aligned}\) 
where we used the Poincaré-Wirtinger inequality to obtain the second inequality. By translation and using the uniform estimates (REF ) and (REF ), we obtain
\(\Vert v^n \Vert _{L_\infty (\mathbb {R}^2)} \le c \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)} \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)},\) 
for some positive constant \(c\) , independent of \(n\) . Additionally, by testing (REF ) with \(v^n \varphi _R\) , where \(\varphi _R\)  is our usual rescaled cut-off function, we obtain that
\(\int _{\mathbb {R}^2} |\nabla v^n|^2 \varphi _R \, \mathrm {d}x = \frac{1}{2} \int _{\mathbb {R}} |v^n|^2 \Delta \varphi _R \, \mathrm {d}x + \int _{\mathbb {R}^2} (\nabla ^\perp \hat{\psi }^n_1 \cdot \nabla \hat{\psi }^n_2) v^n \varphi _R \, \mathrm {d}x.\) 
We thereby deduce from the uniform estimates (REF )-(REF ) that there exists a positive constant \(c\)  independent of \(n\)  such that
\(\Vert \nabla v^n \Vert _{L_2(\mathbb {R}^2)} \le c \Vert \nabla \psi _2 \Vert _{L_\infty (\mathbb {R}^2)} \Vert \nabla \psi _1 \Vert _{L_1(\mathbb {R}^2)}.\) 
By the Banach-Alaoglu theorem, the uniform bounds (REF ) and (REF ) imply the existence of a subsequence (which we still label as \(\lbrace v^n\rbrace _{n\in \mathbb {N}}\) ) and a function \(v\)  such that
\(v^n \overset{*}{\rightharpoonup } v \quad \text{in } L_\infty (\mathbb {R}^2), \qquad \nabla v^n \rightharpoonup \nabla v \quad \text{in } L_2(\mathbb {R}^2).\) 
Let us return now to the equation (REF ). We introduce the following test function \(\zeta \in C^\infty _0(\mathbb {R}^2)\) ; there exists \(R>0\)  such that \(\operatornamewithlimits{supp}\zeta \subset B(R)\) . We find that
\(\int _{\mathbb {R}^2} \nabla v^n \cdot \nabla \zeta \, \mathrm {d}x = \int _{\mathbb {R}^2} (\nabla ^\perp \psi _1 \cdot \nabla \psi _2) \zeta \, \mathrm {d}x\) 
for all \(n \ge 2 R + 1\) . We note that arguing in this way avoids having to directly consider the product of two weakly convergent sequences; on the right-hand side of the above.
Using (REF ) to pass to the limit in the previous equation, we find that
\(\int _{\mathbb {R}^2} \nabla v \cdot \nabla \zeta \, \mathrm {d}x = \int _{\mathbb {R}^2} (\nabla ^\perp \psi _1 \cdot \nabla \psi _2) \zeta \, \mathrm {d}x \qquad \forall \zeta \in C^\infty _0(\mathbb {R}^2),\) 
and hence \(v\)  solves (REF ) in the sense of distributions. Estimate (REF ) follows from the convergences (REF ), estimates (REF )-(REF ) and the weak (resp. weak-*) lower semicontinuity of the norms.


Acknowledgements
N. De Nitti was partially supported by the Alexander von Humboldt foundation and by the TRR-154 project of the Deutsche Forschungsgemeinschaft. F. Hounkpe was funded by the Engineering and Physical Sciences Research Council [EP/L015811/1]. S. Schulz was supported by the Royal Society [RGF/EA/181043]. We thank G. M. Coclite, G. Seregin, and E. Zuazua for helpful conversations on topics related to this work.

